Quartile means separating the given set of data into 4 equal parts by 3 divisions. The three separations are median, lower quartile and upper quartile. The lower quartile is the median value between the first number and its median and the upper quartile is the median value between the median and last number of a given set. The interquartile range outlier means finding the outlier of a given set of data with the help of interquartile range.
Formula involved – Interquartile range outliers:
n – n is the total number of elements in a set.
Median =` (n+1)/2` = `Q_2`
Lowe quartile = `(n+1)/4` = `Q_1`
Upper quartile = `(3(n+1))/4 ` = `Q_3`
Interquartile range = `Q_3-Q_1` (IQR).
The outliers are below `Q_1-1.5*(IQR)` and above `Q_3+1.5*(IQR).`
The inner fences are `Q1-1.5*(IQR)` and `Q3+1.5*(IQR)` and the outer fences are `Q1-3*(IQR)` and `Q3+3*(IQR)` The outliers lies between these fences.
Example problems – Interquartile range outliers:
Example 1 – Interquartile range outliers:
Find the interquartile range outlier for the given set of data 45, 43, 41, 43, 43, 41, 42.
Solution:
The given set of data is 45, 43, 41, 43, 43, 41, 42.
Organize the given set of data in ascending order.
41, 41, 42, 43, 43, 43, 45
The median is `(7+1)/2 ` = `8/2` = 4
The `4^ (th)` value is the median.
Median = 43.
Lower quartile = `(n+1)/4` = `(7+1)/4` = `8/4` = 2.
The element which is in `2^(nd)` position is a lower quartile
41 is the lower quartile.
Upper quartile = `(3(n+1))/4` =`(3(7+1))/4` = 6
The element which is in `6^(th)` position is an upper quartile.
43 is an upper quartile.
Interquartile range = `Q_3-Q_1`
= `43-41`
= `2`
Now to find the outlier it needs to calculate `Q_1-1.5*(IQR)` and `Q_3+1.5*(IQR).`
`Q_1-1.5*(IQR)` = `41-(1.5*2)` = `38`
`Q_3+1.5*(IQR)` = `43 + (1.5 * 2)` = `46`
The outliers are below 38 and above 46.
There is `no` outlier for a given set of data.
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Example 2 – Interquartile range outliers:
Find the interquartile range outlier for the given set of data 10, 21, 25, 24, 27, 45, 30.
Solution:
The given set of data is 10, 21, 25, 24, 27, 45, 30.
Organize the given set of data in ascending order.
10, 21, 24, 25, 27, 30, 45.
The median is `(7+1)/2` = `8/2` =`4`
The `4^(th)` element is a median.
Median = 25.
Lower quartile = `(n+1)/4` = `(7+1)/4` = `2` .
The `2^(nd)` element is a lower quartile.
Lower quartile = 21
Upper quartile =` (3(n+1))/4` =`(3(7+1))/4` = `6`
The `6^(th)` element is an upper quartile.
Upper quartile = 30
Interquartile range = `Q_3-Q_1`
= `30-21`
= `9` .
Now to find the outlier it needs to calculate `Q_1-1.5*(IQR)` and `Q_3+1.5*(IQR).`
`Q_1-1.5*(IQR) ` = `21-(1.5*9)` = `7.5`
`Q_3+1.5*(IQR)` = `30 + (1.5 * 9)` = `43.5`
The outliers are below 7.5 and above 43.5.
So `45` is an outlier of a given set of data.
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